If the symbols $\hat A,\hat B$ mean some functionals of $a,a^\dagger$ which don't depend on $\rho$, your problem has no solution. You simply cannot rewrite objects such as $\rho H$ in the form $H' \rho$; matrices and operators don't commute with each other and their refusal to commute isn't a formality – it's a fundamental fact.
A proof that it can't be done?
The right hand side of your desired equation has the form $X\rho$ where $X=u_1(t)A+u_2(t)B$. Imagine that at $t=0$, there exists a state $|\psi\rangle$ that is annihilated by $\rho$ i.e.
$$\rho |\psi\rangle = 0$$
Then your desired form of the equation implies
$$\dot \rho |\psi\rangle = X\rho|\psi\rangle = 0$$
as well, regardless of the choice of $X$. However, it's clear that the right – first – equation you wrote at the top allows $\dot\rho |\psi\rangle$ to be nonzero. In fact, it will generically be nonzero because states such as $a|\psi\rangle$ and $a^\dagger \psi\rangle$ that appear as the "right side factors" in various terms are generically nonzero and the extra $\rho$ in front of them no longer annihilates them.
I am just saying that if a particular state $|\psi\rangle$ has a vanishing probability to be realized at $t=0$ according to the density matrix, it doesn't mean that it will have a vanishing probability at all later times. However, your desired form of the equation would imply that it stays zero forever. It can't be right so your equation cannot be equivalent to the original one regardless of the choice of $X$.